Problem: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $26.8$ years; the standard deviation is $3.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living between $20.4$ and $30$ years.
Explanation: $26.8$ $23.6$ $30$ $20.4$ $33.2$ $17.2$ $36.4$ $95\%$ $68\%$ $13.5\%$ $13.5\%$ We know the lifespans are normally distributed with an average lifespan of $26.8$ years. We know the standard deviation is $3.2$ years, so one standard deviation below the mean is $23.6$ years and one standard deviation above the mean is $30$ years. Two standard deviations below the mean is $20.4$ years and two standard deviations above the mean is $33.2$ years. Three standard deviations below the mean is $17.2$ years and three standard deviations above the mean is $36.4$ years. We are interested in the probability of a zebra living between $20.4$ and $30$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the zebras will have lifespans within 2 standard deviations of the average lifespan. It also tells us that $68\%$ of the zebras will have lifespans within 1 standard deviation of the mean. The probability of a particular zebra living between $20.4$ and $30$ years is $\color{orange}{13.5\%} + {68\%}$, or $81.5\%$.